Problem: Solve for $x$ : $ 7|x + 9| + 4 = 1|x + 9| + 6 $
Subtract $ {1|x + 9|} $ from both sides: $ \begin{eqnarray} 7|x + 9| + 4 &=& 1|x + 9| + 6 \\ \\ { - 1|x + 9|} && { - 1|x + 9|} \\ \\ 6|x + 9| + 4 &=& 6 \end{eqnarray} $ Subtract ${4}$ from both sides: $ \begin{eqnarray} 6|x + 9| + 4 &=& 6 \\ \\ { - 4} &=& { - 4} \\ \\ 6|x + 9| &=& 2 \end{eqnarray} $ Divide both sides by ${6}$ $ \dfrac{6|x + 9|} {{6}} = \dfrac{2} {{6}} $ Simplify: $ |x + 9| = \dfrac{1}{3}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x + 9 = -\dfrac{1}{3} $ or $ x + 9 = \dfrac{1}{3} $ Solve for the solution where $x + 9$ is negative: $ x + 9 = -\dfrac{1}{3} $ Subtract ${9}$ from both sides: $ \begin{eqnarray} x + 9 &=& -\dfrac{1}{3} \\ \\ {- 9} && {- 9} \\ \\ x &=& -\dfrac{1}{3} - 9 \end{eqnarray} $ Change the ${ - 9}$ to an equivalent fraction with a denominator of $3$ $ x = - \dfrac{1}{3} {- \dfrac{27}{3}} $ $ x = -\dfrac{28}{3} $ Then calculate the solution where $x + 9$ is positive: $ x + 9 = \dfrac{1}{3} $ Subtract ${9}$ from both sides: $ \begin{eqnarray} x + 9 &=& \dfrac{1}{3} \\ \\ {- 9} && {- 9} \\ \\ x &=& \dfrac{1}{3} - 9 \end{eqnarray} $ Change the ${ - 9}$ to an equivalent fraction with a denominator of $3$ $ x = \dfrac{1}{3} {- \dfrac{27}{3}} $ $ x = -\dfrac{26}{3} $ Thus, the correct answer is $x = -\dfrac{28}{3} $ or $x = -\dfrac{26}{3} $.